## Area of Semicircle Found From Lines at Right Angles

Complete the triangle ABC. From C draw a line to the end of the dimater at D. Triangle ACD is a right angled triangle.

If

\[\angle BAC =x\]

then \[\angle BCA = 90-x\]

and \[\angle DCB=90-(90-x)=x, \: \angle BDC=90-x\]

.Triangles ABC and BDC have three identical angles, so are similar triangles.

Then

\[\frac{AB}{BC}=\frac{BC}{BD} \rightarrow BD=\frac{BC \times BC}{AB}=\frac{16^2}{8}=32\]